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Indirect Maximizing

  • Thread starter PFStudent
  • Start date
170
0
1. Homework Statement

Given,
[tex]
{{(x - 7)}^{2}} + {{(y - 3)}^{2}} = {{8}^{2}}
[/tex]

What is,
[tex]
max(3x+4y)
[/tex]

2. Homework Equations

None really.

3. The Attempt at a Solution

Letting,
[tex]
3x+4y = C
[/tex]

When I get to the point where I have,
[tex]
{y} = {{\frac{-3x}{4}}+{\frac{C}{4}}}
[/tex]

Then substitute that in to,
[tex]
{{(x-7)}^{2}} + {{(y-3)}^{2}} = {{8}^{2}}
[/tex]

I get,
[tex]
{{\left(x - 7\right)}^{2}} + {{\left({\left({{\frac { - 3x}{4}} + {\frac {C}{4}}}\right)} - 3\right)}^{2}} = {{8}^{2}}
[/tex]

However, I am not sure how to proceed from here since I have two unknowns: [itex]x[/itex] and [tex]C[/tex].

So, how do I proceed from here?

Thanks,

-PFStudent
 
Last edited:

Answers and Replies

351
2
Can you lagrange?
If I am thinking right, then this is asking you the max on
z = 3x+4y when this intersects (x-7)^2 .. equation
 
351
2
If you go by your way
I would suggest to differentiate the final equation with respect to C (you are maximizing C)
And you will find y = Ax+C equation (with some numbers)
And now, you know it should agree with your (x-7)^2+(y-.. equation
This should work.

Use Lagrange if you know it. It's lot faster and easier
 
170
0
Hey,

If you go by your way
I would suggest to differentiate the final equation with respect to C (you are maximizing C)
And you will find y = Ax+C equation (with some numbers)
And now, you know it should agree with your (x-7)^2+(y-.. equation
This should work.

Use Lagrange if you know it. It's lot faster and easier
Well the way I am interpreting this problem is that in the equation,

[tex]
max(3x+4y) = max(C)
[/tex]

Where [tex]max(C)[/tex] is a constant.

Additionally, if I differentiate as follows,

[tex]
{\frac{d}{dC}{\left[}}{{\left(x - 7\right)}^{2}} + {{\left({\left({{\frac { - 3x}{4}} + {\frac {C}{4}}}\right)} - 3\right)}^{2}}{\right]} = {\frac{d}{dC}{\left[}}{{8}^{2}}{\right]}
[/tex]

How am I supposed to differentiate: implicitly or partially with respect to [tex]C[/tex]?.

In addition, taking the derivative and setting it equal to zero will only yield the values that maximize the original function--however I still do not see how this will find, max(3x+4y).

Thanks,

-PFStudent
 
351
2
partially: treat x as constant.
so you will get C = something*x+some numbers
now substitute C in 3x+4y = C equation
and you will be some line
So, now find intersection of this line with original function(would give u max/min)

I think max(3x+4y) means you take x and y value from your function domain. So, finding function max when x and y are in 3x+4y relationship should give u the answer...or something like that
 
351
2
my interpretation:

Draw a cylinder in x-y-z co-od with that is defined by (x-7)^2 .. equation when z = 0
Draw a plane define by z=3x+4y

you will get a slanted disk, and they are asking for max of that disk
 
1,703
5
parameterize the constraint then plug that parameterization into your function. then maximize subject to the parameterization. the easieast way to parameterize your constraints is x = f(y) then you'll have to check max's on two parameterizations. if you still can't get it i'll post more.

actually just use lagrange multipliers.
 
Last edited:

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